In triangle ABC, the medians AD,BE, and CF concur at the centroid G.
(a) Prove that AD < (AB + AC)/2.
(b) Let P=AB+AC+BC be the perimeter of triangle ABC. Prove that 3P/4 < AD + BE + CF < P.
To prove the given statements, we can use the properties of the centroid and the triangle inequalities.
(a) Proving AD < (AB + AC)/2:
1. Draw line segment AD to link vertex A to the midpoint D of side BC.
2. By the properties of medians, we know that AD divides side BC into two equal segments, BD and DC.
3. Let x = BD and y = DC. Hence, BC = BD + DC = x + y.
4. By the triangle inequality, for any triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Thus, we have AB + AC > BC.
5. Substituting x + y for BC, we get AB + AC > x + y.
6. Since AD is a median, it is less than or equal to half the sum of the lengths of the other two sides. Therefore, AD ≤ (x + y)/2.
7. Combining steps 5 and 6, we have AB + AC > (x + y)/2.
8. Multiply both sides of the inequality by 2: 2(AB + AC) > x + y.
9. Since x + y = BC, we can substitute it back in: 2(AB + AC) > BC.
10. Dividing both sides of the inequality by 2, we obtain: AB + AC > BC/2.
11. Rearranging the inequality, we have AB + AC > AD.
Therefore, AD < (AB + AC)/2 is proven.
(b) Proving 3P/4 < AD + BE + CF < P:
1. The sum of the lengths of the medians of a triangle is equal to three times the length of the centroid to any vertex.
2. Since AD, BE, and CF are medians, we have AD + BE + CF = 3 * AG, where AG is the length from the centroid G to any vertex.
3. Note that AG is the median, which is less than or equal to half the sum of the lengths of the other two sides. Therefore, AG ≤ (AB + AC)/2.
4. Multiply both sides by 3: 3 * AG ≤ 3 * (AB + AC)/2.
5. Simplifying the inequality, we obtain: AD + BE + CF ≤ (3/2) * (AB + AC).
6. Since the perimeter P of triangle ABC is equal to AB + AC + BC, we have AB + AC = P - BC.
7. Substituting P - BC for AB + AC, we get: AD + BE + CF ≤ (3/2) * (P - BC).
8. Rearranging the inequality, we have: AD + BE + CF + (3/2) * BC ≤ (3/2) * P.
9. Since BC is a side of the triangle, it is greater than zero. Therefore, (3/2) * BC > 0.
10. Adding (3/2) * BC to both sides of the inequality, we have: AD + BE + CF + (3/2) * BC < (3/2) * P + (3/2) * BC.
11. Simplifying the right side of the inequality, we obtain: AD + BE + CF + (3/2) * BC < (3/2)(P + BC).
12. Since BC = P - AB - AC, we can substitute it back in: AD + BE + CF + (3/2) * (P - AB - AC) < (3/2)(P + (P - AB - AC)).
13. Simplifying further, we get: AD + BE + CF + (3/2) * P - (3/2) * AB - (3/2) * AC < (3/2) * P + (3/2) * P - (3/2) * AB - (3/2) * AC.
14. Combining like terms, we simplify: AD + BE + CF < (3/2) * P + (3/2) * P.
15. Rearranging the inequality, we have: AD + BE + CF < 3P.
Therefore, we have proven that 3P/4 < AD + BE + CF < P.